If someone offers you even odds on a 10 dollar bet that in six rolls of a die at least one 6 would be rolled. What would the payouts have to be in order to make the expected value of the equal to 0? Is this a good bet to take?

Thanks!

The probability of rolling at least one 6 among six consecutive die rolls is the probability of rolling a 6 first, plus the probability of rolling not a 6 and then a 6, plus the probability of rolling two not-6s and then a 6, etc.

\(\displaystyle P = \frac{1}{6} + \left(\frac{5}{6}\right)\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^3\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^5\left(\frac{1}{6}\right)\)

So basically the $10 is irrelevant because even money just means, if you lose then you lose your bet, and if you win then you gain twice your bet, which can be represented as -1 and 2.

So we have \(\displaystyle E(X) = (2)(P) + (-1)(1-P)\)

where X is winnings (for a bet of $1, or 1 of any unit of currency you choose), a discrete random variable.

If the number you obtain for E(X) is greater than 0, then this is a good bet.

In order to find what the payout would have to be to get E(X) = 0, you must set E(X) = 0 and solve the following equation for m.

\(\displaystyle E(X) = (m)(P) + (-1)(1-P)\)

(I just chose the letter m arbitrarily.)

I usually work with probabilities rather than odds. I'll leave it to you to express m as payout. Even money is 2 for 1, or 1 to 1. So you'll have m for 1, but it seems you should convert this to something

*to* 1, not

*for* 1. If I were a gambler maybe I'd see the point of using odds; probabilities seem so much easier to work with.

Edit: @awkward: as soon as I read "Here is a short-cut" I realized the much easier way. Thanks.